3 Forces on a Ring

1. Jan 28, 2012

zaper

Three forces are applied to a ring (as shown in the photo) that lies on a frictionless surface in the xy plane. The ring has a mass of 100 kg. Fa=200N, Fc=240N and the angle between Fa and Fb is 135°.

What is Fb if:

The system is stationary?

The system accelerates at .5 m/s2?

For some reason I just can't get a grasp on this problem. I understand that to be stationary all forces must cancel out, but I can't figure out how with Fc having a stronger pull in the x direction than Fa how Fb (which appears to go straight down) can stop Fc.

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• phys ring prob.jpg
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Last edited: Jan 28, 2012
2. Jan 28, 2012

tiny-tim

hi zaper!
you aren't told the angle between Fc and the other two …

so you can make it anything you like!

3. Jan 28, 2012

zaper

So Basically to not move I should make the angle between Fb and Fc 135 as well which means that Fa and Fc will cancel x-wise and Fb will have to be 2*Fa*cos(45)?

4. Jan 28, 2012

tiny-tim

no, because Fa is 200 N and Fc is 240 N

5. Jan 28, 2012

zaper

Oh yeah. So then 200cos(45)=240cos(x) so Fc is at 53.9. This means Fb=200sin(45)+240sin(53.9)which is 335.3

Last edited: Jan 28, 2012
6. Jan 28, 2012

tiny-tim

no, not 45°, you need to use an unknown θ

7. Jan 28, 2012

zaper

Yeah I edited my previous post so hopefully it's correct. I'm sorry. My brain is not working this morning

8. Jan 28, 2012

tiny-tim

ahh!
yes! (i haven't checked the figures, but …) that method looks fine

9. Jan 28, 2012

zaper

Ok so that solves the first part. Now for the second part since it goes .5m/s2 in the x direction and the ring is 100 kg then Fc is 50 N greater than Fa in the x direction so 200cos(45)=240cos(x)-50?

10. Jan 28, 2012

zaper

If this method is correct then I get that Fc is at 37.1. This means that Fb=200sin(45)+240sin(37.1) which is 286.2

11. Jan 28, 2012

tiny-tim

yes, but you'll need a y equation also

12. Jan 28, 2012

zaper

I have that in my last post I believe

13. Jan 28, 2012

tiny-tim

ahh!
yes, that looks fine too